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To solve the separable differential equation ૨૨ dy y² - 6 dx we must find two separate integrals: dy = help (formulas) and 10 dx help (formulas) = 1 2y The first integral we integrate by substitution: u = du = help (formulas) help (formulas) Solving for y we get one positive solution y = ☐ help (formulas) And one negative solution y = ☐ help (formulas) Note: You must simplify all arbitrary constants down to one constant k. Find the particular solution satisfying the initial condition y(1) = −5. y(x) = ☐ help (formulas) Book: Section 1.3 of Notes on Diffy Qs

To solve the separable differential equation ૨૨ dy y² - 6 dx we must find two separate integrals: dy = help (formulas) and 10 dx help (formulas) = 1 2y The first integral we integrate by substitution: u = du = help (formulas) help (formulas) Solving for y we get one positive solution y = ☐ help (formulas) And one negative solution y = ☐ help (formulas) Note: You must simplify all arbitrary constants down to one constant k. Find the particular solution satisfying the initial condition y(1) = −5. y(x) = ☐ help (formulas) Book: Section 1.3 of Notes on Diffy Qs

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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To solve the separable differential equation ૨૨ dy y² - 6 dx we must find two separate integrals: dy = help (formulas) and 10 dx help (formulas) = 1 2y The first integral we integrate by substitution: u = du = help (formulas) help (formulas) Solving for y we get one positive solution y = ☐ help (formulas) And one negative solution y = ☐ help (formulas) Note: You must simplify all arbitrary constants down to one constant k. Find the particular solution satisfying the initial condition y(1) = −5. y(x) = ☐ help (formulas) Book: Section 1.3 of Notes on Diffy Qs
Transcribed Image Text:To solve the separable differential equation ૨૨ dy y² - 6 dx we must find two separate integrals: dy = help (formulas) and 10 dx help (formulas) = 1 2y The first integral we integrate by substitution: u = du = help (formulas) help (formulas) Solving for y we get one positive solution y = ☐ help (formulas) And one negative solution y = ☐ help (formulas) Note: You must simplify all arbitrary constants down to one constant k. Find the particular solution satisfying the initial condition y(1) = −5. y(x) = ☐ help (formulas) Book: Section 1.3 of Notes on Diffy Qs
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